By Proposition REF , it is enough to show that the necessity. Assume that \(T\)  is a cosilting module in \(\mathsf {Mod}\hspace{0.72229pt}R\) . By Proposition REF , \(T\)  induces a torsion pair (\(\mathsf {Ker}\hspace{0.72229pt}\mbox{Hom}_{R}(-,T)\) , \(\mathsf {Cogen}\hspace{0.72229pt}T\) ) in \(\mathsf {Mod}\hspace{0.72229pt}R\) . Let \(I\)  be the torsion part of \(R\)  with respect to the torsion pair (\(\mathsf {Ker}\hspace{0.72229pt}\mbox{Hom}_{R}(-,T)\) , \(\mathsf {Cogen}\hspace{0.72229pt}T\) ). We check that this \(I\)  is just the desired right ideal of \(R\) . It is easy to see that \(\mbox{Hom}_{R}(I,\mathcal {B}_{\zeta })=0\) . Then for any \(X\in \mathsf {Cogen}\hspace{0.72229pt}T\) , we have an isomorphism \(\mbox{Hom}_{R}(R/I,X)\cong \mbox{Hom}_{R}(R,X)\) . Thus, \(X\)  is generated by \(R/I\)  and so \(\mathsf {Cogen}\hspace{0.72229pt}T\subseteq \sigma [R/I]\) . Note that \(R/I\in \mathsf {Cogen}\hspace{0.72229pt}T=\mathcal {B}_{\zeta }\subseteq {^{\perp }T}\) . Since \(\mathsf {Ext}^{1}_{\sigma [R/I]}(R/I,T)\subseteq \mathsf {Ext}^{1}_{R}(R/I,T)=0\) , we have \(\mathsf {Ext}^{1}_{\sigma [R/I]}(R/I,T)=0\) .
Now, it is enough to check (1). Since \(T\in \mathcal {B}_{\zeta }\) , we know that \(\mbox{Hom}_{R}(I,T)=0\) . By Lemma REF , there exists a short exact sequence in \(\sigma [R/I]\)
\(0\longrightarrow T\longrightarrow \mathsf {Tr}_{R/I}(Q_{0})\xrightarrow{}\mathsf {Tr}_{R/I}(Q_{1})\longrightarrow 0.\) 
Then the injective dimension of \(T\)  in \(\sigma [R/I]\)  is at most one.
Next, we prove that \(\mathsf {Ext}^{1}_{\sigma [R/I]}(T^{\lambda }, T)=0\) . Note that \(T\in \mathcal {B}_{\zeta }\subseteq {^{\perp }T}\)  and \(\mathcal {B}_{\zeta }\)  is closed under product. Thus for any set \(\lambda \) , \(T^{\lambda }\in {^{\perp }T}\) . Moreover, since \(\mathsf {Cogen}\hspace{0.72229pt}T\subseteq \sigma [R/I]\) , \(T^{\lambda }\in \sigma [R/I]\)  and so \(\mathsf {Ext}^{1}_{\sigma [R/I]}(T^{\lambda }, T)\subseteq \mathsf {Ext}^{1}_{R}(T^{\lambda }, T)=0\) . It means that \(\mathsf {Ext}^{1}_{\sigma [R/I]}(T^{\lambda }, T)=0\) .
From [1]}, there exists a short exact sequence in \(\sigma [R/I]\)
\(0\rightarrow T_{1}\rightarrow T_{0}\rightarrow \mathsf {Tr}_{R/I}(E) \rightarrow 0\) 
where \(T_{i}\in \mathsf {Prod}\hspace{0.72229pt}T\) , \(E\)  is an injective cogenerator of \(\mathsf {Mod}\hspace{0.72229pt}R\) . By Lemma REF , we know that \(T_{i}\in \mathsf {Prod}\hspace{0.72229pt}_{\sigma [R/I]}T\) , where \(\mathsf {Prod}\hspace{0.72229pt}_{\sigma [R/I]}T\)  consists of all direct summands of product of \(T\)  in \(\sigma [R/I]\) .
We claim that \(\mathsf {Ker}\hspace{0.72229pt}\mbox{Hom}_{\sigma [R/I]}(-,T)\bigcap \mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)=0\) .
For any \(X\in \mathsf {Ker}\hspace{0.72229pt}\mbox{Hom}_{\sigma [R/I]}(-,T)\bigcap \mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)\) , applying \(\mbox{Hom}_{\sigma [R/I]}(X,-)\)  to the sequence (REF ), we have the following exact sequence
\(\mbox{Hom}_{\sigma [R/I]}(X,T_{0})\rightarrow \mbox{Hom}_{\sigma [R/I]}(X,\mathsf {Tr}_{R/I}(E))\rightarrow \mathsf {Ext}^{1}_{\sigma [R/I]}(X,T_{1}).\) 
Since \(\mbox{Hom}_{\sigma [R/I]}(X,T^{\lambda })=\mbox{Hom}_{R}(X,T)^{\lambda }=\mbox{Hom}_{\sigma [R/I]}(X,T)^{\lambda }\) , we have that \(\mbox{Hom}_{\sigma [R/I]}(X,T_{0})=0\) . By Lemma REF  and \(X\in \mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)\) , \(\mathsf {Ext}^{1}_{\sigma [R/I]}(X,T_{1})=0\) . Hence, we have that \(\mbox{Hom}_{\sigma [R/I]}(X,\mathsf {Tr}_{R/I}(E))=0\) . Since \(\mathsf {Tr}_{R/I}(E)\)  is an injective cogenerator, there is a monomorphism \(X\rightarrow \prod _{\alpha \in J}^{R/I}\mathsf {Tr}_{R/I}(E)\) . Then, by Lemma REF , we have the monomorphism \(g\)  which is the composition of the following monomorphisms
\(\mbox{Hom}_{\sigma [R/I]}(X,X)\rightarrow \mbox{Hom}_{\sigma [R/I]}(X,\Pi _{\alpha \in J}^{R/I}\mathsf {Tr}_{R/I}(E)_{\alpha })\rightarrow \Pi _{\alpha \in J}\mbox{Hom}_{\sigma [R/I]}(X,\mathsf {Tr}_{R/I}(E)_{\alpha }).\) 
Then we obtain that \(\mathsf {End}_{R}(X)=0\)  and so, \(X=0\) .
Now, we check that \(\mathsf {Cogen}\hspace{0.72229pt}_{\sigma [R/I]}T=\mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)\) . For any \(X\in \mathsf {Cogen}\hspace{0.72229pt}_{\sigma [R/I]}T\) , there is a monomorphism \(f:X\rightarrow \prod _{\alpha \in J}^{R/I}T_{\alpha }\) . By Lemma REF , we have that \(\prod _{\alpha \in J}T_{\alpha }=T^{J}\) . Hence, \(X\in \mathsf {Cogen}\hspace{0.72229pt}T=\mathcal {B}_{\zeta }\subseteq {^{\perp }T}\) . It implies that \(X\in \mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)\)  and hence, \(\mathsf {Cogen}\hspace{0.72229pt}_{\sigma [R/I]}T\subseteq \mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)\) . On the other hand, for any \(X\in \mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)\) , there is a short exact sequence in \(\sigma [R/I]\)
\(0\rightarrow \mathsf {Rej}_{T}(X)\rightarrow X\rightarrow X/\mathsf {Rej}_{T}(X)\rightarrow 0.\) 
Applying \(\mbox{Hom}_{\sigma [R/I]}(-,T)\)  to this sequence, we have a long exact sequence
\(0\rightarrow &\mbox{Hom}_{\sigma [R/I]}(X/\mathsf {Rej}_{T}(X),T) {\cong }\mbox{Hom}_{\sigma [R/I]}(X,T) \rightarrow \mbox{Hom}_{\sigma [R/I]}(\mathsf {Rej}_{T}(X),T)\\\rightarrow &\mathsf {Ext}^{1}_{\sigma [R/I]}(X/\mathsf {Rej}_{T}(X),T)\rightarrow \mathsf {Ext}^{1}_{\sigma [R/I]}(X,T)\rightarrow \mathsf {Ext}^{1}_{\sigma [R/I]}(\mathsf {Rej}_{T}(X),T)\\\rightarrow & \mathsf {Ext}^{2}_{\sigma [R/I]}(X/\mathsf {Rej}_{T}(X),T)\rightarrow \cdots .\) 

By the formula in [2]} and \(\sigma [R/I]\)  is a full subcategory, we have the isomorphism \(\mbox{Hom}_{\sigma [R/I]}(X/\mathsf {Rej}_{T}(X),T) \cong \mbox{Hom}_{\sigma [R/I]}(X,T)\) . Since \(X/\mathsf {Rej}_{T}(X)\)  is cogenerated by \(T\) , from Lemma REF , \(X/\mathsf {Rej}_{T}(X)\in \mathsf {Cogen}\hspace{0.72229pt}_{\sigma [R/I]}T\) . Then, \(\mathsf {Ext}^{1}_{\sigma [R/I]}(X/\mathsf {Rej}_{T}(X),T)=0\) . It yields that \(\mbox{Hom}_{\sigma [R/I]}(\mathsf {Rej}_{T}(X),T)=0\) . Since \(X\in \mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)\)  and the injective dimension of \(T\)  in \(\sigma [R/I]\)  is at most one, we know that \(\mathsf {Ext}^{1}_{\sigma [R/I]}(X,T)=0=\mathsf {Ext}^{2}_{\sigma [R/I]}(X/\mathsf {Rej}_{T}(X),T)\) . Thus, \(\mathsf {Ext}^{1}_{\sigma [R/I]}(\mathsf {Rej}_{T}(X),T)=0\) . This means \(\mathsf {Rej}_{T}(X)\in \mathsf {Ker}\hspace{0.72229pt}\mbox{Hom}_{\sigma [R/I]}(-,T)\bigcap \mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)\) . Hence, \(\mathsf {Rej}_{T}(X)=0\)  and so, \(X\cong X/\mathsf {Rej}_{T}(X)\in \mathsf {Cogen}\hspace{0.72229pt}_{\sigma [R/I]}T\) . Therefore, \(\mathsf {Ker}\hspace{0.72229pt}\mathsf {Ext}^{1}_{\sigma [R/I]}(-,T)\subseteq \mathsf {Cogen}\hspace{0.72229pt}_{\sigma [R/I]}T\) . This completes the proof.
